Hi all,

IS this correct?

thanks

John

[math}\int \frac{x.dx}{x^2+4x+13}[/tex]

Printable View

- Sep 10th 2011, 02:32 AMceltic1234Integration partial fractions
Hi all,

IS this correct?

thanks

John

[math}\int \frac{x.dx}{x^2+4x+13}[/tex] - Sep 10th 2011, 02:38 AMe^(i*pi)Re: Integration partial fractions
You've only stated the question? Partial fractions is a good method here but first some algebra:

$\displaystyle \int \dfrac{10}{(2x-6)(x+5)} = \int \dfrac{10}{2(x-3)(x+5)} = 5 \int \dfrac{1}{(x-3)(x+5)}$

$\displaystyle \dfrac{1}{(x-3)(x+5)} = \dfrac{A}{x-3} + \dfrac{B}{x+5} \Leftrightarrow 1 = A(x+5) + B(x-3)$

Can you continue?

edit: That other integral is not an easy one to work out. Are you familiar with trig integrals? - Sep 10th 2011, 02:42 AMceltic1234Re: Integration partial fractions
sorry i am busy in the background trying to get laTex to work-i'll try and sort it ASAP

- Sep 10th 2011, 02:43 AMe^(i*pi)Re: Integration partial fractions
- Sep 10th 2011, 02:45 AMceltic1234Re: Integration partial fractions
I am solving the partial fraction to give A= 5/4 and B=-1/6.

The integral solves as:

$\displaystyle \frac{1}{4}5ln(2x-6) - \frac{1}{6}ln(x+1) +c $

Can you tell me if i am correct?

thanks

John - Sep 10th 2011, 02:55 AMe^(i*pi)Re: Integration partial fractions
I get B = -1/8

Sub in x=-5: $\displaystyle 1 = B(-5-3) = -8B \Leftrightarrow B = -\dfrac{1}{8}$

Sub in x=3: $\displaystyle 1 = 8A \Leftrightarrow A = \dfrac{1}{8}$

Hence $\displaystyle 5 \int \dfrac{1}{(x-3)(x+5)} = 5 \left(\int \dfrac{1}{8(x-3)} - \int \dfrac{1}{8(x+5)}\right) = \dfrac{5}{8} \left(\int \dfrac{1}{x-3} - \int \dfrac{1}{x+5}\right)$

$\displaystyle =\dfrac{5}{8} \left( \ln(x-3) - \ln(x+5) + C_1\right) = \dfrac{5}{8} \ln \left(\dfrac{x-3}{x+5}\right) + C \text{ where } C = \dfrac{5}{8}C_1$

Perhaps it would be easier to make a new thread since there seem to be two questions on the go here! Alternatively I can delete my posts regarding the original integral - Sep 10th 2011, 02:58 AMceltic1234Re: Integration partial fractions
ok thanks

i'll do a new thread